Need help on this word problem..?

Question

A gas company needs to mix 2 different octane levels of gasoline to achieve the ideal mixture required by the customer. 500 gallons of 89% octane gasoline is obtained by mixing 87% octane gasoline and 90% octane gasoline. How much of each must be used?

Can anyone please help? Anything in the right direction would be much appreciated. Thanks

Answer 1

We've got two different equations here with two variables. They break down like this:

Let X = Amount (gal) of 87% Octane Gas in mixture
Let Y = Amount (gal) of 90% Octane Gas in mixture

The two things we know from the word problem are:

87%x + 90%y = 89%(500)
x + y = 500

Take the second equation, and solve for y:

y = 500 - x

then, substitute your value for Y into the first equation, and solve for X:

0.87x + .90(500 - x) = .89(500)
0.87x + 450 - .90x = 445
0.87x - 0.90x = 445 - 450
-.03x = -5
x = 166.666

Now, solving for Y is easy once we know X

Y = 500 - X
Y = 500 - 166.666
Y = 333.333

So, there is 166 and two-thirds gallons of 87% Octane Gas and 333 and one-thirds gallons of 90% Octane Gas.

Answer 2

You need 2 parts 90% and 1 par 87%. That means altogether you will have three parts. 500/3= 166 2/3

90% = 2*166 2/3 = 333 1/3 gallons of 90%

87% = 166 2/3 gallons of 87%

Altogether, it comes out to 500 gallons of 89%.

Answer 3

Okay, letting x be how many gallons of 87% octane, and y be how many gallons of 90% octane, then x+y=500 , and .87x+.9y=.89*500.
can you solve it from there?

Answer 4

x+y=500
87x+90y=500*89
87x+87y=500*87
subtracting
3y=500*2
y=1000/3
=333 1/3 gallons of 87% and
166 2/3 gallons of 90%

Answer 5

let's say you need x gallons of 87% octane gassoline and y gallons of 90%

x*87+ y*90=(x+y)*89
87x +90y=89x+89y
y=2x

=> you need twice the amount of the 90% one than the 87%one

Answer 6

490 gallons of the 87% plus 10 gallons of the 90%.

Answer 7

let x = gallons of 87%
then the # of gallons of 90% is (500 - x)

Answer 8

x = # gallons at 87%
500-x = # of gallons at 90 %
So, .87x + .9 (500-x) = 500(.89)
.87x + 450 -.9x = 445
-.03x = -5
x = 166 2/3 gallons

Answer 9

You need to mix 2wice as much 90 as 87...

Answer 10

C1V1=C2V2 <---best formula ever